an aircraft factory manufactures airplane engines. The unit cost c (the cost in dollars to make each airplane engine) depends on the number of engines made. If x engines are made, then the unit cost is given by the function c(x)=0.3x2- 96x+26,919. How many engines must be made to minimize the unit cost?
The vertex is the minimum or maximum point of a parabola.
The formula to find the vertex is, where a, b and c are values in the standard form of quadratic equation is:
x min/max = - b / 2a
If a > 0, then the parabola the y-value of the vertex is the minimum value of the function.
If a < 0, then the parabola the y-value of the vertex is the maximum value of the function.
The formula to find the vertex is, where a, b and c are values in the standard form of quadratic equation is:
x min/max = - b / 2a
If a > 0, then the parabola the y-value of the vertex is the minimum value of the function.
If a < 0, then the parabola the y-value of the vertex is the maximum value of the function.
In your case:
c(x) = 0.3 x² - 96 x + 26 919
a = 0.3 , b = - 93 , c = 26 919
a > 0 , so the y-value of the vertex is the minimum value of the function.
x min = - b / 2a = - ( - 96 ) / 2 ∙ 0.3 = 96 / 0.6 = 160
160 engines must be made to minimize the unit cost.
c min = c ( 160 ) = 0.3 ∙ 160² - 96 ∙ 160 + 26 919 = 19 239
My typo.
b is not - 93
b = - 96
a = 0.3 , b = - 96 , c = 26 919
To minimize the unit cost, we need to find the value of x that corresponds to the minimum point on the cost function.
The cost function is given by c(x) = 0.3x^2 - 96x + 26,919.
To find the minimum point, we need to differentiate the cost function and set the derivative equal to zero.
Taking the derivative of c(x) with respect to x, we get:
c'(x) = 0.6x - 96
Now, let's set c'(x) = 0 and solve for x:
0.6x - 96 = 0
0.6x = 96
x = 96 / 0.6
x = 160
So, to minimize the unit cost, the factory must make 160 engines.
To minimize the unit cost, we need to find the value of x that corresponds to the minimum point on the cost function c(x)=0.3x^2-96x+26,919.
The cost function represents a quadratic equation with a positive coefficient for the x^2 term (0.3x^2). Since the coefficient is positive, the graph of the quadratic will be an upward-opening parabola.
To find the minimum point of the quadratic function c(x), we can use calculus. We take the derivative of c(x) with respect to x, set it equal to zero, and solve for x.
So, let's find the derivative of c(x):
c'(x) = 0.3(2x) - 96
= 0.6x - 96
Setting c'(x) equal to zero and solving for x:
0.6x - 96 = 0
0.6x = 96
x = 96 / 0.6
x = 160
Now, we have the value of x that corresponds to the critical point. But we need to verify if it is a minimum point. To do that, we can take the second derivative of c(x) and check its sign.
Taking the second derivative of c(x):
c''(x) = 0.6
Since the second derivative is positive (0.6), this confirms that x = 160 corresponds to a minimum point on the cost function.
Therefore, 160 engines must be made to minimize the unit cost.